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(a) In the set of numbers from 75, 76, 77, ... , 497, 498, 499, 500, you will find that the first term which is the first number divisible by 7 = 77 and the last term = 497. Given the first and last terms of an A.P, you can use the formula \(n = \frac{l - a}{d} + 1\) to find the number of terms in the sequence. First term = a = 77 Last term = l = 497 Common difference = d = 7 \(n = \frac{497 - 77}{7} + 1\) \(n = \frac{420}{7} + 1\) = \(60 + 1\) = 61. There are 61 numbers between 75 and 500 which are divisible by 7. (b) \(T_{n} = a + (n - 1)d\) (for an A.P) \(T_{8} = a + 7d\) \(T_{3} = a + 2d\) \(\implies a + 7d = 5(a + 2d)\) \(T_{7} = a + 6d\) \(T_{4} = a + 3d\) \(\implies a + 6d = a + 3d + 9\) \(a + 7d = 5a + 10d \implies 5a - a + 10d - 7d = 0\) \(4a + 3d = 0 ... (1)\) \(a + 6d = a + 3d + 9 \implies a - a + 6d - 3d = 9\) \(3d = 9 \implies d = 3\) Putting d = 3 in (1), \(4a + 3(3) = 0 \implies 4a + 9 = 0\) \(4a = -9 \implies a = \frac{-9}{4}\) \(\therefore\) The second term of the sequence = \(\frac{-9}{4} + 3 = \frac{3}{4}\) Third term = \(\frac{3}{4} + 3 = \frac{15}{4}\) Fourth term = \(\frac{15}{4} + 3 = \frac{27}{4}\) Fifth term = \(\frac{27}{4} + 3 = \frac{39}{4}\) The first 5 terms of the sequence = \(\frac{-9}{4}, \frac{3}{4}, \frac{15}{4}, \frac{27}{4}, \frac{39}{4}\)

1101\(_2\) + 10111\(_2\) + 111\(_2\) = 101011\(_2\)

125\(_x\) = 20\(_{10}\) 1 x X\(^2\) + 2 x X\(^1\) + 5x X \(^0\) = 20 X2 + 2X + 5 = 20 X\(^2\) + 2X - 15 = 0 (X + 5)( X - 3) = 0 X + 5 implies X = -5 X - 3 implies X = 3 But X cannot be negative ∴X = 3

No expanded working has been attached for this question yet. The verified answer is Option D.

NB: This is a correction to the solution in the video lesson. 123456 \(\approxeq\) 123000.

\(I = \frac{P \times T \times R}{100}\) = \(\frac{7500 \times 8 \times 5}{100}\) = N3000

N85 - N60 = N25 increase ∴percentage increase = \(\frac{25}{60}\times \frac{100}{1}\) = \(\frac{125}{3}\) = 41.67% = 42%

\(16^{\frac{-1}{2}}\times 4^{\frac{-1}{2}} \times 27^{\frac{1}{3}}\) =\(\frac{1}{16^{\frac{1}{2}}}\times \frac{1}{4^{\frac{1}{2}}}\times 27^{\frac{1}{3}}\) =\(\frac{1}{\sqrt{16}} \times \frac{1}{\sqrt{4}} \times \sqrt[3]{27}\) =\(\frac{1}{4} \times \frac{1}{2} \times \frac{3}{1}\) = \(\frac{3}{8}\)

If log\(_{x^{\frac{1}{2}}}\)64 = 3 (X\(^{\frac{1}{2}})^3\) = 64 (X\(^{\frac{1}{2}})^3\) = 4\(^3\) X\(^{\frac{1}{2}}\) = 4 X = 4\(^2\) X = 16

\(\frac{1+\sqrt{2}}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}}\) = \(\frac{1+(1+\sqrt{2})+\sqrt{2}(1+\sqrt{2})}{1^2 - (\sqrt{2})^2}\) = \(\frac{(1+\sqrt{2}+\sqrt{2}+2)}{1-2}\) = \(\frac{3+2\sqrt{2}}{-1}\) = - 3 - 2\(\sqrt{2}\) ∴ X and Y = -3 and -2

X = {1,5,10} Y = {3,4,6} X∩Y = ∅

(4x+3y)\(^2\) - (3x-2y)\(^2\) Using the differences of two squares: (4x+3y+3x-2y)(4x+3y-(3x-2y)) (4x+3y+3x-2y)(4x+3y-3x+2y) (x+5y)(7x+y)

n(M) only = 30-10 = 20 n(E) only = 20-10 = 10 n(M∩E) = 10 ∴M∪E = 20+10+10 = 40

\(L=\frac{4}{3}M\sqrt{PQ}\\ =\frac{3}{4M} \times L = \sqrt{PQ}\\ =\left(\frac{3L}{4M}\right)^2=(\sqrt{PQ})^2\\ =\frac{9L^2}{16M^2}=PQ\\ =Q=\frac{9L^2}{16M^2 P}\)

2x2 - kx - 12 is divisible by x-4 implies x is a factor ∴ x = 4 f(4) implies 2(4)2 - k(4) - 12 = 0 32 - 4k - 12 = 0 -4k + 20 = 0 -4k = -20 k = 5